# Quadrilateral inscribed in a circle

• May 5th 2011, 10:56 PM
gundanium
Quadrilateral inscribed in a circle
Let a quadrilateral ABCD be inscribed in a circle such that AB=5, BC=3, CD=2 and Angle B=60 degrees.
Find length AC
Find length DA

Okay, i found AC but i can't seem to find DA. I know im suppose to relate the 60 degrees to some other triangle but I just can't seem to find any relation. Am I missing a postulate about internal angles of a quadrilateral inscribed in a circle? XD
• May 6th 2011, 12:49 AM
earboth
Quote:

Originally Posted by gundanium
Let a quadrilateral ABCD be inscribed in a circle such that AB=5, BC=3, CD=2 and Angle B=60 degrees.
Find length AC
Find length DA

Okay, i found AC but i can't seem to find DA. I know im suppose to relate the 60 degrees to some other triangle but I just can't seem to find any relation. Am I missing a postulate about internal angles of a quadrilateral inscribed in a circle? XD

I have a result, but I hope for you that there is a more simple and elegant way to do the question. Here is my method:

1. Use triangle ABC to calculate the length of the radius:
$r = \dfrac{\frac12 \overline{AC}}{\sin(60^\circ)}$

2. You'll get 4 isosceles triangles with the radius as their equal legs. Calculate the interior angles of the 3 known isosceles triangles at the center of the circle. the 4th angle add up to 360°. I've got $\delta \approx 73.174^\circ$

3. Calculate the length of
$\overline{AD} = 2 \cdot r \cdot \sin\left(\frac12 \delta\right)$

I've got AD = 3.
• May 6th 2011, 05:42 AM
bjhopper
This problem can be solved by coordinate geometry not as simple as using trig but certainly more elegant.It appears that the quadrilateral is a regular trapezoid and AB a chord close to the center of the circumscribed circle